Characterizing Near-Constant Functions in Discrete Product Spaces

[WannaBe22]

[Topology] Product Spaces :(

Homework Statement

1. Show that in the product space N^N where the topology on N is discrete, the set of near-constant functions is dense (near constant function is a function that becomes constant from a specific index..)...

2. Prove that in R^I the set of monotonic increasing functions is not open.

Homework Equations

The Attempt at a Solution

I've no idea how to start thinking of these questions...

I'll be delighted to receive some guidance

Thanks in advance

 

[rasmhop]

1) A set is dense if its closure is the whole space. That is, if for every point f \in \mathbb{N}^\mathbb{N} every neighborhood of f intersects the set, but it's sufficient to consider all subbasis elements containing f since all other open neighborhoods can be formed with intersections and unions of these. Let D denote the set of near-constant functions and consider an element f \in \mathbb{N}^\mathbb{N}. Consider a subbasis element U that contains f. Its nth component is \mathbb{N} except for at one index k. Now define,
g(x) = \begin{cases} f(x) & \textrm{if }x = k \\ 0 & \textrm{otherwise} \end{cases}
Then, g is in D and in the neighborhood U of f.

2) If the set of monotonic increasing functions were open then, then every monotonically increasing function would have an open neighborhood that consists only of monotonic increasing functions. However if U is such a neighborhood of a monotonically increasing function f, then U is \mathbb{R} at all points except for finitely many. Thus you can choose a function g in U that is not monotonically increasing by choosing a sufficiently small value at one of the points where U is \mathbb{R}. For instance if U is \mathbb{R} at m > 0, then consider,
g(x) = \begin{cases} f(0)-1 & \textrm{if }x=m \\ f(x) & \textrm{otherwise} \end{cases}

 

[WannaBe22]

Thanks a lot! You're very helpful!